Question

The volume of a spherical balloon is increasing at the rate of 25cm3/sec. Find the rate of change of its surface area at the instant when the radius is 5 cm.

Answer 1

Answer:

$10cm^2/s$

Step-by-step explanation:

The volume of a spherical balloon is given by:

$V = \frac{4}{3} \pi r^3$

Differentiating with respect to time, we get:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ ...1

It is given that the rate of change of volume is $25cm^3/s$ and the radius is 5cm, so:

$25 = 4\pi (5)^2 \frac{dr}{dt}$

Or, $\frac{dr}{dt} = \frac{1}{4\pi} cm/s$

Now, the surface area of the balloon is given by:

$S = 4\pi r^2$

Differentiating, again, with respect to time:

$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

We now know the values of r and $\frac{dr}{dt}$ both, so:

$\frac{dS}{dt} = 8\pi * 5 * \frac{1}{4\pi} cm^2/s$

So, $\frac{dS}{dt} = 10cm^2/s$

Or, the rate of change of its surface area when the radius is 5cm, is equal to $10cm^2/s$.

Answer 2

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Let V be volume of sphere of radius r.

$V = (\frac{4}{3}) × πr³$

From given condition,

$\frac{dV}{dt} = 25 \:\:cm³/s$

:. $\frac{d}{dt} × (\frac{4}{3}πr³)$ = $25\:\: cm³/s$

$=> \frac{4}{3}π × 3r.\frac{dr}{dt} = 25 cm³/s$

$\frac{dr}{dt}$ = $\frac{25}{(4πr²)}$ — (1)

:. Surface area = $4πr²$

$\frac{d (surface\:\:area)}{dt} = \frac{d}{dt}(4πr²)$

$=> 8πr \frac{dr}{dt} = 8πr × (\frac{25}{4πr²}$

$= \frac{50}{r}$

When r =5, rate of increase of surface area

$=> \frac{50}{5}$

$=> 10 \:\:cm²/s$

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